Introduction
A power series is an infinite series of the form:
\sum_{n=0}^{\infty} c_n (x - a)^n
where
c_n
are coefficients,
a
is the center of the series, and
x
is the variable.
The series converges for certain values of
x
and diverges for others.
The range around
a
where the series converges is the Radius of Convergence,
R
:
\begin{align*}
\text{If } |x - a| &< R, \text{ the series converges.} \\
\text{If } |x - a| &> R, \text{ the series diverges.}
\end{align*}
Lastly, we must check if the series converges or diverges when
|x-a| = R
Then, we can define the Interval of Convergence, which is either:
\begin{align*}
&[a - R, a + R], \\
&(a - R, a + R), \\
&[a - R, a + R), \text{ or} \\
&(a - R, a + R]
\end{align*}
Examples
The radius of convergence is almost always found using the Ratio Test or Root Test. The first three examples use the Ratio Test, and the last three use the Root Test.
Example 1: Ratio Test
Find the radius and interval of convergence of
\sum_{n=1}^{\infty} \frac{x^n}{n}
.
\begin{align*}
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left| \frac{x^{n+1}}{n+1} \cdot \frac{n}{x^n} \right| \\
&= |x| \lim_{n \to \infty} \frac{n}{n+1} \\
&= |x| \cdot 1 = |x|
\end{align*}
For convergence, we need
|x| < 1
, so the radius of convergence is
R = 1
.
Now we check the endpoints:
\begin{align*}
x = 1: \quad &\sum_{n=1}^{\infty} \frac{1^n}{n} = \sum_{n=1}^{\infty} \frac{1}{n} \quad \text{Diverges (Harmonic Series)} \\
x = -1: \quad &\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \quad \text{Converges (Alternating Harmonic Series)}
\end{align*}
The interval of convergence is
[-1, 1)
.
Example 2: Ratio Test
Find the radius and interval of convergence of
\sum_{n=0}^{\infty} \frac{2^n x^n}{n!}
.
\begin{align*}
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left| \frac{2^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n x^n} \right| \\
&= \lim_{n \to \infty} \left| \frac{2x}{n+1} \right| \\
&= |2x| \cdot 0 = 0
\end{align*}
Since the limit is
0 < 1
for all
x
, the series converges for all
x
.
The radius of convergence is
R = \infty
, and the interval of convergence is
(-\infty, \infty)
.
Example 3: Ratio Test
Find the radius and interval of convergence of
\sum_{n=1}^{\infty} n! x^n
.
\begin{align*}
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left| \frac{(n+1)! x^{n+1}}{n! x^n} \right| \\
&= \lim_{n \to \infty} \left| (n+1)x \right| \\
&= |x| \lim_{n \to \infty} (n+1) \\
&= \begin{cases} 0 & \text{if } x = 0 \\ \infty & \text{if } x \neq 0 \end{cases}
\end{align*}
The limit is
\infty > 1
for
x \neq 0
, so the series diverges for all
x \neq 0
.
The series only converges when
x = 0
.
The radius of convergence is
R = 0
, and the interval of convergence is
\{0\}
.
Example 1: Root Test
Find the radius and interval of convergence of
\sum_{n=1}^{\infty} \left(\frac{2x+1}{3}\right)^n
.
\begin{align*}
\lim_{n \to \infty} \sqrt[n]{|a_n|} &= \lim_{n \to \infty} \sqrt[n]{\left|\left(\frac{2x+1}{3}\right)^n\right|} \\
&= \lim_{n \to \infty} \left|\frac{2x+1}{3}\right| \\
&= \left|\frac{2x+1}{3}\right|
\end{align*}
For convergence, we need
\left|\frac{2x+1}{3}\right| < 1
, which simplifies to
|2x+1| < 3
or
-2 < x < 1
.
The radius of convergence is
R = \frac{3}{2}
.
Now we check the endpoints:
\begin{align*}
x = 1: \quad &\sum_{n=1}^{\infty} \left(\frac{2(1)+1}{3}\right)^n = \sum_{n=1}^{\infty} 1^n \quad \text{Diverges} \\
x = -2: \quad &\sum_{n=1}^{\infty} \left(\frac{2(-2)+1}{3}\right)^n = \sum_{n=1}^{\infty} (-1)^n \quad \text{Diverges}
\end{align*}
The interval of convergence is
(-2, 1)
.
Example 2: Root Test
Find the radius and interval of convergence of
\sum_{n=1}^{\infty} \frac{(x+3)^{2n}}{n}
.
\begin{align*}
\lim_{n \to \infty} \sqrt[n]{|a_n|} &= \lim_{n \to \infty} \sqrt[n]{\left|\frac{(x+3)^{2n}}{n}\right|} \\
&= \lim_{n \to \infty} \frac{|x+3|^2}{\sqrt[n]{n}} \\
&= |x+3|^2 \lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} \\
&= |x+3|^2 \cdot 1 = |x+3|^2
\end{align*}
For convergence, we need
|x+3|^2 < 1
, which simplifies to
|x+3| < 1
or
-4 < x < -2
.
The radius of convergence is
R = 1
.
Now we check the endpoints:
\begin{align*}
x = -2: \quad &\sum_{n=1}^{\infty} \frac{(-2+3)^{2n}}{n} = \sum_{n=1}^{\infty} \frac{1}{n} \quad \text{Diverges (Harmonic Series)} \\
x = -4: \quad &\sum_{n=1}^{\infty} \frac{(-4+3)^{2n}}{n} = \sum_{n=1}^{\infty} \frac{1}{n} \quad \text{Diverges (Harmonic Series)}
\end{align*}
The interval of convergence is
(-4, -2)
.
Example 3: Root Test
Find the radius and interval of convergence of
\sum_{n=1}^{\infty} \frac{(x-5)^n}{5^n n^5}
.
\begin{align*}
\lim_{n \to \infty} \sqrt[n]{|a_n|} &= \lim_{n \to \infty} \sqrt[n]{\left|\frac{(x-5)^n}{5^n n^5}\right|} \\
&= \lim_{n \to \infty} \frac{|x-5|}{5 \sqrt[n]{n^5}} \\
&= \frac{|x-5|}{5} \lim_{n \to \infty} \frac{1}{(\sqrt[n]{n})^5} \\
&= \frac{|x-5|}{5} \cdot \frac{1}{1^5} = \frac{|x-5|}{5}
\end{align*}
For convergence, we need
\frac{|x-5|}{5} < 1
, which simplifies to
|x-5| < 5
or
0 < x < 10
.
The radius of convergence is
R = 5
.
Now we check the endpoints:
\begin{align*}
x = 10: \quad &\sum_{n=1}^{\infty} \frac{(10-5)^n}{5^n n^5} = \sum_{n=1}^{\infty} \frac{1}{n^5} \quad \text{Converges (p-series, p=5>1)} \\
x = 0: \quad &\sum_{n=1}^{\infty} \frac{(0-5)^n}{5^n n^5} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^5} \quad \text{Converges (Alternating Series)}
\end{align*}
The interval of convergence is
[0, 10]
.